a) A+B=\(x^2-2xy+y^2\)+\(y^2+2xy+5\)
=\(x^2\)+(\(-2xy+2xy\))+\(\left(y^2+y^2\right)\)+5
=\(x^2\)+\(2y^2\)+5
A-B=(\(x^2-2xy+y^2\))-(\(y^2+2xy+5\))
=\(x^2-2xy+y^2-y^2-2xy-5\)
=\(x^2+\left(-2xy-2xy\right)+\left(y^2-y^2\right)\)-5
=\(x^2\)\(-4xy-5\)
b) Ta có \(\dfrac{1}{5}xy^2z\left(-5xy\right)\)=\(\left[\dfrac{1}{5}.\left(-5\right)\right].\left(x.x\right).\left(y^2.y\right).z\)
=-1.\(x^2.y^3.z\)
Phần hệ số -1
Phần biến: \(x^2.y^3.z\)
a) A+B=x2−2xy+y2x2−2xy+y2+y2+2xy+5y2+2xy+5
=x2x2+(−2xy+2xy−2xy+2xy)+(y2+y2)(y2+y2)+5
=x2x2+2y22y2+5
A-B=(x2−2xy+y2x2−2xy+y2)-(y2+2xy+5y2+2xy+5)
=x2−2xy+y2−y2−2xy−5x2−2xy+y2−y2−2xy−5
=x2+(−2xy−2xy)+(y2−y2)x2+(−2xy−2xy)+(y2−y2)-5
=x2x2−4xy−5−4xy−5
b) Ta có [15.(−5)].(x.x).(y2.y).z[15.(−5)].(x.x).(y2.y).z
=-1.x2.y3.zx2.y3.z
Phần hệ số -1
Phần biến: x2.y3.z
